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Avva
 2 years ago
Best ResponseYou've already chosen the best response.0would you plz write the function using equation

amistre64
 2 years ago
Best ResponseYou've already chosen the best response.0hmm, something to do with:\[f_{xy}f_{yx}(f_{xx})^2\] or did i recall that incorrectly?

amistre64
 2 years ago
Best ResponseYou've already chosen the best response.0almost had it :) \[f_{xx}f_{yy}(f_{xy})^2\]

Living_dreams
 2 years ago
Best ResponseYou've already chosen the best response.1\[4+x ^{3}+y ^{3}3xy\]

Living_dreams
 2 years ago
Best ResponseYou've already chosen the best response.1it fxx.fyy(fxy)^2

Living_dreams
 2 years ago
Best ResponseYou've already chosen the best response.1i jst want to verify the answer for it !

TuringTest
 2 years ago
Best ResponseYou've already chosen the best response.0what answer did you get?

Avva
 2 years ago
Best ResponseYou've already chosen the best response.0\[fx = 2x3y \] \[fy= 2y3x\] on solving both equations the point is (0,0) \[fxx = 2\] , \[fyy= 2\] \[fxy = \] \[D = 5 <0 saddle point\]

TuringTest
 2 years ago
Best ResponseYou've already chosen the best response.0that's not quite right I don't think

TuringTest
 2 years ago
Best ResponseYou've already chosen the best response.0\[f(x,y)=4+x^3+y^33xy\]\[f_x=3x^23y=0\implies y=x^2\]\[f_y=3y^23x=3x^43x=3x^3(x1)=0\]\[ x=\{0,1\},y=\{0,1\}\]

Living_dreams
 2 years ago
Best ResponseYou've already chosen the best response.1its x^3 so i have (0,0) ,(1,1)and a (1,1) i am not sure if (1,1) should be there oor not

Avva
 2 years ago
Best ResponseYou've already chosen the best response.0@Living_dreams I solved what you mentioned in your comment

TuringTest
 2 years ago
Best ResponseYou've already chosen the best response.0there is no x=1 solution, I don't see how you get that

TuringTest
 2 years ago
Best ResponseYou've already chosen the best response.0@Avva I don't see how you get your fx and fy

Avva
 2 years ago
Best ResponseYou've already chosen the best response.0Ooh I saw the powers on X ^ y 2 not 3

Living_dreams
 2 years ago
Best ResponseYou've already chosen the best response.1@Avva thanks for ur help but i was wonder abt the max and min points! n\ @TuringTest : 3y^23x=0 i sub y=x^2 from other equation so it will be 3x(x^31)=0 therefore you have x =0,1,1 i am not sure abt 1 though !! tht was my question

TuringTest
 2 years ago
Best ResponseYou've already chosen the best response.0x^3=1 does not have x=1 as a solution since (1)^3=1, not 1

TuringTest
 2 years ago
Best ResponseYou've already chosen the best response.0I did screw up my factoring earlier though, you were right about that :P

TuringTest
 2 years ago
Best ResponseYou've already chosen the best response.03x(x^31)=0 3x=0 > x=0 x^3=1 > x=1

Living_dreams
 2 years ago
Best ResponseYou've already chosen the best response.1@TuringTest that alrite!! :) so now my question is x^3=1 so will i hv x=1 if i solve for x ?

TuringTest
 2 years ago
Best ResponseYou've already chosen the best response.0no, again, because \((1)^3\neq1\) so \(x^3=1\) has only one real solution, x=1 x=1 is NOT a solution

Avva
 2 years ago
Best ResponseYou've already chosen the best response.0the other two roots are complex not real

Living_dreams
 2 years ago
Best ResponseYou've already chosen the best response.1@TuringTest oh ok thanks a lot for ur help!! so the final ans will be (0,0 ) is a saddle point and (1,1) will be a local minimum !

Living_dreams
 2 years ago
Best ResponseYou've already chosen the best response.1@Avva :thanks for ur help !! :)

Avva
 2 years ago
Best ResponseYou've already chosen the best response.0@Living_dreams URW :) sorry I didn't get that the powers are 3 at first

Living_dreams
 2 years ago
Best ResponseYou've already chosen the best response.1it okay !!i knw it get hard to tell the difference between 2 and 3 if we type it in equation form !!
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